NCERT Class 10 Mathematics Chapter 13 Statistics

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NCERT Class 10 Mathematics Chapter 13 Statistics

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Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per (CBSE) Book guidelines. NCERT Class 10 Mathematics Textual Question Answer. These solutions are part of NCERT All Subject Solutions. Here we have given NCERT Class 10 Mathematics Chapter 13 Statistics Solutions for All Subject, You can practice these here.

Statistics

Chapter – 13

Exercise 13.1

1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

Number of plants 0 – 22 – 44 – 6 6 – 8 8 – 10 10 – 1212 – 14
Number of houses 1234567

Which method did you use for finding the mean, and why?

Ans: To find the class mark (xi) for each interval, the following relation is used.

Class mark 𝒙1 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕+𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/ 2

Number of plantsNumber of houses (fi) xifixi 
0 − 2111 × 1 = 1
2 − 4232 × 3 = 6
4 − 6151 × 5 = 5
6 − 8575 × 7 = 35
8 − 10696 × 9 = 54
10 − 12212 ×11 = 22
12 − 143133 × 13 = 39
Total20162

From the table, it can be observed that

Therefore, mean number of plants per house is 8.1. 

2. Consider the following distribution of daily wages of 50 worker of a factory.

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Daily wages (in Rs)100 − 120120 − 140140 −1 60160 − 180180 − 200
Number of workers 12148610

Ans: To find the class mark for each interval, the following relation is used.

X1 = upper limit + Lower Limit// 2

Class size (h) of this data  = 20 

Taking 150 as assured mean (a) , d1, u1 and f1u1 can be calculate as follows 

Daily wages (in Rs)Number of workers  (fi) xidi = xi − 150𝒖𝒊 = 𝒅𝒊 /𝟐0fiui
100 −12012110− 40-2-24
120 − 14014130− 20-114
140 − 1608150000
160 −18061702016
180 − 2001019040220
Total50-12

From the table, it can be observed that

Therefore, the mean daily wage of the workers of the factory is Rs 145.20

3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs.18. Find the missing frequency f.

Daily pocket allowance (in Rs) 11−1313−1515−1717−1919−2121−2323−25
Number of workers76913f54

Ans: To find the class mark (xi) for each interval, the following relation is used:

X1 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/2

Given that, mean pocket allowance,

Taking 18 as assured mean (a), di and f1d1 are calculated as follows.

Daily pocket allowance (in Rs)Number of children fiClass mark xidi = xi − 18 fidi 
11 −13712-6-42
13 − 15614-4-24
15 − 17916-2-18
17 −19131800
19 − 21f2022f
21 − 23522420
23 − 25 4424624
Total2f-40

Hence, the missing frequency, f, is 20.

4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Fine the mean heart beats per minute for these women, choosing a suitable method.

Number of heart beats per minute65−6868−7171−7474−7777−8080−8383−8
Number of women2438742

Ans: To find the class mark of each interval (xi), the following relation is used.

X1 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊 / 2 

Class size, h, of this data = 3 

Taking 75.5 as assumed mean (a) assumed (a) ui, f1u1 are calculated as follows.

Number of heart beats per minuteNumber of women fixidi = xi − 75.5𝒖𝒊 = 𝒅𝒊/3fiui
65 − 68266.5− 9-3-6
68 − 71469.5− 6-2-8
71 − 74369.5− 3-1-3
74 − 77875.5000
77 − 80778.5337
80 − 83481.5 6668
83 − 86284.5996
Total304

Therefore, mean hear beats per minute for these women are 75.9 beats per minute. 

5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

Number of mangoes50 − 5253 − 55 56−5859−6162−64
Number of boxes 1511013511525

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Ans: 

Number of mangoes Number of boxes fi
50 − 5215
53 − 55 110
56 − 58135
59 − 61115
62 − 6425

It can be observed that class intervals are not continuous. There is a gap of 1 between two class intervals. Therefore, 1/2 has to be added to the upper class limit.

and 1/2 has to be subtracted from the lower class limit of each interval.

Class mark (xi) can be obtained by using the following relation.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size (h) of this data = 3

Taking 57 as assumed mean (a), di, ui, f1u1are calculated as follows

Class intervalfixidi = xi − 57𝒖𝒊 = 𝒅i/3f1xu1
49.5 − 52.51551-6-2-30
52.5 − 55.511054-3-1-110
55.5 − 58.513557000
58.5 − 61.51156031115
61.5 − 64.525636250
Total 4400

Mean number of mangoes kept in a packing box is 57.19.

Step deviation method is used here as the values of fi, di are big and also, there is a common multiple between all di.

6. The table below shows the daily expenditure on food of 25 households in a locality.

Daily expenditure (in Rs)100−150150−200200−250250−300300−350
Number of households451223

Find the mean daily expenditure on food by a suitable method. 

Ans: To find the class mark (xi) for each interval, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size = 50

Taking 225 as assumed mean (a), di, ui, f1u1are calculated as follows. 

Daily expenditure (in Rs)f1xidi = xi − 225𝒖𝒊 = 𝒅𝒊 50 ff1u1
100 − 1504125-100-2-8
100 − 1505175-50-1-5
200 − 25012225000
250 − 30022755012
300 − 350232510024
Total25

Therefore, mean daily expenditure on food is Rs 211. 

7. To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below: 

concentration of SO2 (in ppm)Frequency
0.00 − 0.044
0.04 − 0.089
0.08 − 0.129
0.12 − 0.162
0.16 − 0.204
0.20 − 0.242

Ans: To find the class marks for each interval, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊/2 

lass size of this data = 0.04 

Taking 0.14 as assumed mean (a), di, ui, f1u1are calculated as follows. 

Concentration of SO2 (in ppm)Frequency fClass mark xidi = xi − 0.14𝒖𝒊 = 𝒅𝒊 0.04fiui
0.00 − 0.0440.02-0.12− 3− 12
0.04 − 0.0890.06-0.08− 2− 18
0.08 − 0.1290.10-0.04− 1− 9
0.12 − 0.1620.14000
0.16 − 0.2040.18114
0.20 − 0.24 20.22224
Total 30−31

Therefore, mean concentration of SO2 in the air is 0.099 ppm. 

8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent. 

Number of days0−66−1010−1414−2020−2820−2838−40
Number of students111074431

Ans: To find the class mark of each interval, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊/2

Taking 17 as assumed mean (a), di and f1d1 are calculated as follows.

Number of daysNumber of studentsxidi = xi − 17fid
0 − 6113− 14− 154
6 − 10108− 9− 90
10 − 14712− 5− 35
14 − 204400
20 − 28424728
28 − 383331648
38 − 401392222 
Total40− 181

From the table, we obtain

Therefore, the mean number of days is 12.48 days for which a student was absent. 

9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

Literacy rate (in %) 45 − 5555 − 6565 − 7575 − 85 85 − 95
Number of cities3101183

Ans: To find the class marks, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Class size (h) for this data = 10

Taking 70 as assumed mean (a), di, ui, and fiui are calculated as follows.

Literacy rate (in %)Number of cities fix1di = x1 − 70 𝒖𝒊 = 𝒅𝒊 1f1 u1 
45 − 55350-20-2-6
55 − 651060-10-10-10
65 − 751170-000
75 − 858801018
85 − 953902026
Total35− 2

From the table, we obtain

Therefore, mean literacy rate is 69.43%. 

Exercise 13.2

1. The following table shows the ages of the patients admitted in a hospital during a year: 

Age (in years)5−1515 − 25 25−3535−4545−5555−65
Number of patients6112123 145

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Ans: To find the class marks (xi), the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Taking 30 as assumed mean (a), di and f1d1are calculated as follows. 

Age (in years)Number of patients fiClass mark xidi = xi − 30fidi
5 − 15610-20-120
15 − 251120– 10-110
25 − 35213000
35 − 45234010230
45 − 55145020280
55 − 6556030150
Total80430

Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years.

It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45.

Modal class = 35 − 45

Lower limit (l) of modal class = 35

Frequency (f1) of modal class = 23

Class size (h) = 10

Frequency (f0) of class preceding the modal class = 21

Frequency (f2) of class succeeding the modal class = 14 

2.  Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years. 

Lifetimes (in hours) 0 − 2020−4040−6060−8080−100100−120
Frequency 103552613829

Determine the modal lifetimes of the components. 

3. The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure. 

Expenditure (in Rs)Number of families 
1000 − 150024
1500 − 2000 40
2000 − 250033
2500 − 300028
3000 − 3500 30
3500 − 4000 22
4000 − 450016
4500 − 5000 7

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. 

Number of students per teacherNumber of states/U.T
15 − 203
20 − 258
25 − 309
30 − 3510
35 − 403
40 − 450
45 − 500
50 − 552

Ans: It can be observed from the given data that the maximum class frequency is 10

belonging to class interval 30 − 35.

Therefore, modal class = 30 − 35

Class size (h) = 5

Lower limit (l) of modal class = 30

Frequency (f1) of modal class = 10

Frequency (f0) of class preceding modal class = 9

Frequency (f2) of class succeeding modal class = 3 

Mode = 30.6

It represents that most of the states/U.T have a teacher-student ratio as 30.6.

To find the class marks, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/2

Taking 32.5 as assumed mean (a), di, ui, and fiui are calculated as follows.

Number of students per teacherNumber of states/U.T (fi)xidi = xi − 32.5𝒖𝒊 = 𝒅𝒊/5f1u1
15 − 20317.5− 15-3− 9
20 − 25822.5− 10-2−16
25 − 30927.5− 5-1− 9
30 − 351032.5000
35 − 40337.5513
40 − 45042.51020
45 − 50 047.51530
50 − 55252.52048
Total35−23

Therefore, mean of the data is 29.2.

It represents that on an average, teacher−student ratio was 29.2.

5. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scoredNumber of batsmen
3000 − 40004
4000 − 500018
5000 − 60009
6000 − 70007
7000 − 80006
8000 − 90003
9000 − 100001
10000 − 110001

Find the mode of the data.

Ans: From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000−5000. Therefore, modal class = 4000−5000

Lower limit (l) of modal class = 4000

Frequency (f1) of modal class = 18

Frequency (f0) of class preceding modal class = 4

Frequency (f2) of class succeeding modal class = 9

Class size (h) = 1000

Therefore, mode of the given data is 4608.7 runs  

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data: 

Number of cars 0−1010−2020−3030−4040−5050−6060−7070−80
Frequency71413122011158

Ans: 

Exercise 13.3

1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption (in units)Number of consumers
65 − 854
85 − 1055
105 − 12513
125 − 14520
145 − 16514
165 − 185 8
185 − 2054

Ans: To find the class marks, the following relation is used.

𝒙𝒊 =𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Taking 135 as assumed mean (a), di, ui, fiui are calculated according to step deviation method as follows. 

Monthly consumption (in units)Number of consumers (f i)xi class mark di= xi− 135𝒖𝒊 = 𝒅𝒊 𝟐𝟎 fiui
65 − 85475− 60− 3−12
85 − 105595− 40− 2− 10
105 − 125 113115− 40− 1− 13
125 − 14520135000
145 − 1651415520114
165 − 185817540216
185 − 205419560312
Total 687

From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 − 145.

Modal class = 125 − 145

Lower limit (l) of modal class = 125

Class size (h) = 20

Frequency (f1) of modal class = 20

Frequency (f0) of class preceding modal class = 13

Frequency (f2) of class succeeding the modal class = 14 

(ii) To find the median of the given data, cumulative frequency is calculated as follows.

Monthly consumption (in units)Number of consumersCumulative frequency
65 − 8544
85 − 10554 + 5 = 9
105 − 125139 + 13 = 22
125 − 1452022 + 20 = 42
145 − 1651442 + 14 = 56
165 − 185856 + 8 = 64
185 − 205464 + 4 = 68 

From the table, we obtain n = 68 Cumulative frequency (cf) just greater than n/2 

(i.e,68/2=34) is 42 belonging to interval 125 − 145.

Therefore, median class = 125 − 145

Lower limit (l) of median class = 125

Class size (h) = 20

Frequency (f) of median class = 20

Cumulative frequency (cf) of class preceding median class = 22

Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05

Respectively.

The three measures are approximately the same in this case.

2. If the median of the distribution is given below is 28.5, find the values of x and y.

Class interval: 0-1010 – 2020 – 3030 -4040-50 50-60Total
Frequency: 5 x2015y560

Ans: The cumulative frequency for the given data is calculated as follows.

50 − 60FrequencyCumulative frequency
0 − 1055
10 − 20x5+ x
20 − 302025 + x 
30 − 401540 + x
40 − 50y40+ x + y
50 − 60545 + x + y
Total (n)60

From the table, it can be observed that n = 60

45 + x + y = 60 or x + y = 15 ..……………….(1)

Median of the data is given as 28.5 which lies in interval 20 − 30.

Therefore, median class = 20 − 30

Lower limit (l) of median class = 20

Cumulative frequency (cf) of class preceding the median class = 5 + x

Frequency (f) of median class = 20

Class size (h) = 10 

From equation (1),

8 + y = 15 y = 7

Hence, the values of x and y are 8 and 7 respectively

3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.

Age (in years)Number of policy holders
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 50 92
Below 5598
Below 60100

Ans: Here, class width is not the same. There is no requirement of adjusting the frequencies according to class intervals. The given frequency table is of less than type represented with upper class limits. The policies were given only to persons with age 18 years onwards but less than 60 years. Therefore, class intervals with their respective cumulative frequency can be defined as below.

Age (in years)Number of policy holders (fi)Cumulative frequency (cf)
18 − 2022
20 − 256 − 2 = 46
25 − 3024 − 6 = 1824
30 − 3545 − 24 = 2145
35 − 4078 − 45 = 3378
40 − 4589 − 78 = 1189
45 − 5092 − 89 = 392
50 − 5598 − 92 = 698
55 − 60100 − 98 = 2100
Total (n)

From the table, it can be observed that n = 100.

Cumulative frequency (cf) just greater than is 78, belonging to

interval 35 − 40

Therefore, median class = 35 − 40

Lower limit (l) of median class = 35

Class size (h) = 5

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 45

Therefore, the median age is 35.76 years. 

4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm)Number or leaves f
118 − 1263
127 − 1355
136 − 1449
145 − 15312
154 − 1625
163 − 1714
172 − 1802

Find the median length of the leaves.

Ans:  

Length (in mm)Number or leaves fCumulative frequency
117.5 − 126.533
126.5 − 135.553 + 5 = 8 
135.5 − 144.598 + 9 = 17
144.5 − 153.5 1217 + 12 = 29
153.5 − 162.5529 + 5 = 34
162.5 − 171.5 434 + 4 = 38
171.5 − 180.5238 + 2 = 40

From the table, it can be observed that the cumulative frequency just greater than

is 29, belonging to class interval 144.5 − 153.5.

Median class = 144.5 − 153.5

Lower limit (l) of median class = 144.5

Class size (h) = 9

Frequency (f) of median class = 12

Cumulative frequency (cf) of class preceding median class = 17

Therefore, median length of leaves is 146.75 mm. 

5. Find the following table gives the distribution of the life time of 400 neon lamps:

Life time (in hours)Number of lamps 
1500 − 200014 
2000 − 250056
2500 − 300060
3000 − 350086
3500 − 400074
4000 − 450062
4500 − 500048

Ans: The cumulative frequencies with their respective class intervals are as follows.

Life timNumber of lamps (fi)Cumulative frequency
1500 − 20001414
2000 − 25005614 + 56 = 70
2500 − 30006070 + 60 = 130
3000 − 350086130 + 86 = 216
3500 − 400074 216 + 74 = 290 
4000 − 450062290 + 62 = 352
4500 − 500048352 + 48 = 400
Total (n)400

It can be observed that the cumulative frequency just greater than n/2 (i.e 400/2 = 200)

Median class = 3000 − 3500

Lower limit (l) of median class = 3000

Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130

Class size (h) = 500

= 3406.976

Therefore, median life time of lamps is 3406.98 hours.

6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters1−4 4−77−1010−13 13−1616−19
Number of surnames63040 644

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames. 

Ans: The cumulative frequencies with their respective class intervals are as follows:

Number of lettersFrequency (fi)Cumulative frequency
1 − 466
4 − 73030 + 6 = 36
7 − 104036 + 40 = 76
10 − 131676 + 16 = 92
13 − 16492 + 4 = 96
16 − 19496 + 4 = 100
Total (n) 100

It can be observed that the cumulative frequency just greater than n/2 (i .e , 100/2 = 50)

is 76, belonging to class interval 7 − 10

Median class = 7 − 10

Lower limit (l) of median class = 7

Cumulative frequency (cf) of class preceding median class = 36

Frequency (f) of median class = 40

Class size (h) = 3 

= 8.05

To find the class marks of the given class intervals, the following relation is used.

𝒙𝒊 = 𝑼𝒑𝒑𝒆𝒓 𝒍𝒊𝒎𝒊𝒕 + 𝑳𝒐𝒘𝒆𝒓 𝒍𝒊𝒎𝒊𝒕/𝟐

Taking 11.5 as assumed mean (a), di, ui, and fiui are calculated according to step deviation method as follows.  

(ii) Taking 11.5 as assumed mean (a), di, ui, and fiui are calculated according to step deviation method as follows. 

Ans: 

Number of letters Number of surnames fifi xidi = xi− 11.5𝒖𝒊 = 𝒅𝒊 𝟑 fiui
1 − 462.5− 9− 3− 18
4 − 7305.5− 6− 2− 60
7 − 10408.5− 3− 1 − 40
10 − 131611.5000
13 − 16414.5314
16 − 19417.5628
Total100

From the table, we obtain

Mean, 

= 11.5 − 3.18 = 8.32

The data in the given table can be written as

Number of letters Frequency (fi)
1 − 46
4 − 730
7 − 1040
13 − 1616
13 − 164
16 − 194
100

From the table, it can be observed that the maximum class frequency is 40

belonging to class interval 7 − 10. Modal class = 7 − 10

Lower limit (l) of modal class = 7

Class size (h) = 3

Frequency (f1) of modal class = 40

Frequency (f0) of class preceding the modal class = 30

Frequency (f2) of class succeeding the modal class = 16 

Therefore, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.88. 

7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students. 

Weight (in kg)40−4545−5050−55 55−6060−65 65−7070−75
Number of students2386632

Ans: The cumulative frequencies with their respective class intervals are as follows: 

Weight (in kg)Frequency (fi)Cumulative frequency
40 − 4522
45 − 5032 + 3 = 5
50 − 5585 + 8 = 13
55 − 60613 + 6 = 19
60 − 65619 + 6 = 25
65 − 70325 + 3 = 28
70 − 75228 + 2 = 30
Total (n) 2

Cumulative frequency just greater than is 19, belonging to class

interval 55 − 60.

Median class = 55 − 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) of median class = 13

Class size (h) = 5

= 56.67

Therefore, the median weight is 56.67 kg.

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